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3 years ago

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Tech A says that a gear set that has a drive gear with 9 teeth and a driven gear with 27 teeth has a gear ratio of 3:1. Tech B s

ays that the drive gear is also called the output gear. Who is correct?

1 answer:

choli [55]3 years ago

5 0

Answer:

Tech A is correct.

Explanation:

Gears are toothed wheels that can be used to transmit power. When two or more gears are in tandem, a gear train is formed.

Gear ratio = $\frac{number of teeth of the driven gear}{Number of teeth of the driving gear}$

= $\frac{27}{9}$

= $\frac{3}{1}$

Gear ratio = 3:1

The driver gear is called the input gear since it transfers its power to the driven gear. While the driven gear is called the output gear because it produces an effect due to both gears.

Tech A is correct.

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Answer:

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From Heat Transfer we determine that heat transfer rate due to electromagnetic radiation ( $\dot Q$ ), measured in BTU per hour, is represented by this formula:

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$\sigma$ - Stefan-Boltzmann constant, measured in BTU per hour-square feet-quartic Rankine.

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$T_{b}$ - Temperature of the bus, measured in Rankine.

If we know that $\epsilon = 0.90$ , $A = 16.188\,ft^{2}$ , $\sigma = 1.714\times 10^{-9}\,\frac{BTU}{h\cdot ft^{2}\cdot R^{4}}$ , $T_{s} = 554.07\,R$ and $T_{b} = 527.67\,R$ , then the heat transfer rate due to electromagnetic radiation is:

$\dot Q = (0.90)\cdot (16.188\,ft^{2})\cdot \left(1.714\times 10^{-9}\,\frac{BTU}{h\cdot ft^{2}\cdot R^{4}} \right)\cdot [(554.07\,R)^{4}-(527.67\,R)^{4}]$

$\dot Q = 417.492\,\frac{BTU}{h}$

Under the consideration of steady heat transfer we find that the net energy transfer from the student's body during the 20 min-ride to school is:

$Q = \dot Q \cdot \Delta t$ (2)

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$Q = \left(417.492\,\frac{BTU}{h} \right)\cdot \left(\frac{1}{3}\,h \right)$

$Q = 139.164\,BTU$

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